PCAP Object-Oriented Programming Practice Question

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

• ✗

  class Meta(type): def __new__(cls, name, bases, dct): dct['created_by']='Meta'

  Why it's wrong here

  Because this `__new__` only adds a key to `dct` and does not return the result of `super().__new__(cls, name, bases, dct)`, the metaclass call returns `None`. Python then binds the class name to that `None` result, so the class object is never created and `C.created_by` cannot be accessed on a real class. Mutating the namespace dictionary alone is not enough; a metaclass `__new__` must construct and return the new class object.

• ✗

  class Meta(type): def __init__(cls, name, bases, dct): dct['created_by']='Meta'

  Why it's wrong here

  The `__init__` hook receives the already-built class object, and by that point `type.__new__` has consumed the namespace dictionary to populate the class's attributes. Adding `created_by` to `dct` here only mutates the original namespace dict, not the class itself, so `C.created_by` remains undefined. The correct place to alter the namespace is in `__new__`, before `super().__new__` creates the class.

• ✗

  def Meta(name, bases, dct): dct['created_by']='Meta'; return type(name, bases, dct)

  Why it's wrong here

  Although Python permits an arbitrary callable as the `metaclass=` argument, this is a plain factory function rather than a metaclass class. It works by constructing a new class through `type`, but it bypasses the actual metaclass protocol (`__prepare__`, `__new__`, `__init__`) and cannot be inherited from as a type-based metaclass. The PCAP exam expects a metaclass to be a class that subclasses `type`, not a standalone function that happens to return a class.

• ✓

  class Meta(type): def __new__(cls, name, bases, dct): dct['created_by']='Meta'; return super().__new__(cls, name, bases, dct)

  Why this is correct

  This correctly overrides `type.__new__`, modifies the mutable namespace dictionary before class construction, and delegates to `super().__new__` to build the class object. Because `super().__new__` copies the entries of `dct` into the new class's `__dict__`, the added `created_by` attribute is present on the finished class. The explicit `return` is essential: without it, no class object would be created at all.